Answer to Question #155716 in Analytic Geometry for Md Tariqul Islam

Question

ansform the equation 17𝑥

2 + 18𝑥𝑦 − 7𝑦

2 − 16𝑥 − 32𝑦 − 18 = 0 to one in 

which there is no term involving 𝑥, 𝑦.

Answer 1

$17𝑥^2 + 18𝑥𝑦 − 7𝑦^2 − 16𝑥 − 32𝑦 − 18 = 0$

The rotation angle:

$tan2\theta=\frac{B}{A-C}$

where A, B and C are the corresponding coefficients from the equation:

$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$

$tan2\theta=\frac{18}{1+7}=\frac{3}{4}=\frac{2tan\theta}{1-tan^2\theta}$

$tan\theta=\frac{1}{3}$

The required transformation equations needed to rotate the coordinate axes to eliminate the xy

term in the given equation for the conic section are:

$x=x'cos\theta-y'sin\theta$

$y=x'cos\theta+y'sin\theta$

$cos\theta=3/\sqrt{10}, sin\theta=1/\sqrt{10}$

$x=\frac{3x'-y'}{\sqrt{10}}, y=\frac{x'+3y'}{\sqrt{10}}$

Substituting these expressions for x and y into the original equation for the conic section and simplifying, we get:

$20x^{'2}-10y^{'2}-\frac{80}{\sqrt{10}}x'-\frac{80}{\sqrt{10}}y'-18=0$

Simplifying and completing the squares to obtain an equation for a hyperbola in standard form, we get:

$\frac{(x'-2/\sqrt{10})^2}{1/2}-\frac{(y'+4/\sqrt{10})^2}{1}=1$